A tent forms an equilateral triangular prism, with both triangular faces exposed. Each side of the triangular face has a length
1 answer:
Answer:
Step-by-step explanation:
By the description in the problem of the tent, there are 3 rectangle faces are faces, one is the floor and the other two are formin the ceiling. This means the height of the tent can be calculated from the information of the triangular faces.
Each side of the triangle is 196 cm, if we divide the equilateral triangle into two rights triangles with a base of 196/2=98cm and a hypotenuse of 196 cm, the side missing is the height of the original equilateral triangle and also the height os the tent. Using the Pitagora's theorem:
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Answer:
<h3>- The ratio of the measure of central angle PQR to the measure of the entire circle is One-eighth. </h3><h3>- The area of the shaded sector depends on the length of the radius. </h3><h3>- The area of the shaded sector depends on the area of the circle</h3>Step-by-step explanation:
Given central angle PQR = 45°
Total angle in a circle = 360°
Ratio of the measure of central angle PQR to the measure of the entire circle is . This shows ratio that <u>the measure of central angle PQR to the measure of the entire circle is one-eighth</u>.
Area of a sector =
= central angle (in degree) = 45°
r = radius of the circle = 6
Area of the sector
<u>The ratio of the shaded sector is 4.5πunits² not 4units²</u>
From the formula, it can be seen that the ratio of the central angle to that of the circle is multiplied by area of the circle, this shows <u>that area of the shaded sector depends on the length of the radius and the area of the circle.</u>
Since Area of the circle = πr²
Area of the circle = 36πunits²
The ratio of the area of the shaded sector to the area of the circle =
For length of an arc
ratio of the length of the arc to the area of the circle =
It is therefore seen that the ratio of the area of the shaded sector to the area of the circle IS NOT equal to the ratio of the length of the arc to the area of the circle
Answer:
(1). y = x ~ Exp (1/3).
(2). Check attachment.
(3). EY = 3(1 - e^-2).
(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.
Step-by-step explanation:
Kindly check the attachment to aid in understanding the solution to the question.
So, from the question, we given the following parameters or information or data;
(A). The probability in which attempt to establish a video call via some social media app may fail with = 0.1.
(B). " If connection is established and if no connection failure occurs thereafter, then the duration of a typical video call in minutes is an exponential random variable X with E[X] = 3. "
(C). "due to an unfortunate bug in the app all calls are disconnected after 6 minutes. Let random variable Y denote the overall call duration (i.e., Y = 0 in case of failure to connect, Y = 6 when a call gets disconnected due to the bug, and Y = X otherwise.)."
(1). Hence, for FY(y) = y = x ~ Exp (1/3) for the condition that zero is equal to y = x < 6.
(2). Check attachment.
(3). EY = 3(1 - e^-2).
(4). Var[y] = 3(1 - e^-2) (1 -3 (1 - e^-2)) - 36e^-2.
The condition to follow in order to solve this question is that y = 0 if x ≤ 0, y = x if 0 ≤ x ≤ 6 and y = 6 if x ≥ 6.
